Projection of line on plane

**patzer** · June 1st, 2011, 05:15 PM

Given $\pi: \begin{cases}l_1: \frac{x-2}{4}\ = \frac{y-1}{2}\ = \frac{z+5}{-4} &\\l_2: \frac{x+4}{-2}\ = \frac{y+1}{0}\ = \frac{z}{1} & \end{cases}$ and the point $M=(1,2,3)$ outside the plane. Find the projection of the line $M_1P$ on plane $\pi$ where $P$ is the intersection point of lines $l_1, l_2$ .

All I can do is that I can find the equation of plane $\pi$ but don't have any idea what to do next.

I need your help.

Thank you.

**Plato** · June 2nd, 2011, 03:33 AM

Originally Posted by patzer

Given $\pi: \begin{cases}l_1: \frac{x-2}{4}\ = \frac{y-1}{2}\ = \frac{z+5}{-4} &\\l_2: \frac{x+4}{-2}\ = \frac{y+1}{0}\ = \frac{z}{1} & \end{cases}$ and the point $M=(1,2,3)$ outside the plane. Find the projection of the line $M_1P$ on plane $\pi$ where $P$ is the intersection point of lines $l_1, l_2$ .

First a comment: No competent textbook or instructor would write a like this one did. It should be $l_2: \frac{x+4}{-2}\ = \frac{z}{1};~y=-1.$ .

Now I will not do the work for you.
First show the line intersect in $P$ .
The directions of the lines are $D_1:<4,2,-4>~\&~D_2:<-2,0,1>$ .
The plane containing $P$ and with normal $N=D_1\times D_2$ is $\pi$ .
The projection is the intersection of the plane containing $P$ with normal $\overrightarrow {PM} \times N$ with plane $\pi.$

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